1.
Exercise set 1.3
11: The argument is valid, since whenever the premises are both true, the con-
clusion is true:
premises
pq
T
T
F
F
conclusion
p q q
p
T
F
T
F
T
F
T
T
F
T
F
T
F
F
T
T
*
12:
(b) This argument is not valid, because there is a case where
Valid Argument Forms
pq
if n is divisible by 10, then n is divisible by 5
p
n is divisible by 10
q
therefore n is divisible by 5
Modus tollels
pq
if n is divisible by 10, then n is divisible by 5
q
n is NOT divisible by 5
p
therefore n is NOT divisible by
1. Exercise set 1.1
7: m c.
13:
Here is the truth table:
pq
p q (p q) (p q) (p q)
T
T
T
F
T
T
F
F
F
F
F
T
F
F
F
F
F
F
T
T
15:
Here is the truth table:
p r qr
p (q r)
pq
r
T
T
T
F
F
T
F
T
T
F
F
T
T
F
T
F
T
F
F
F
F
T
F
F
F
T
T
F
F
T
T
T
F
T
T
F
T
F
T
T
T
T
%# !
&$"
cfw_2 cfw_1, 2
cfw_2 cfw_1, cfw_2 , cfw_3
cfw_1 cfw_1, cfw_2
1 cfw_1
1 cfw_1
1 cfw_1 , 2
7
32
)(
@
6
1
'
:
:
:
:
1 < x 1
x 1 3 x
x 1 1 < x
x < 2 3 x
cfw_x R
cfw_x R
cfw_x R
cfw_x R
2 x < 3
x < 2 1 < x
x < 2 3 x
x 1 1 < x
:
:
:
:
7
54
1
)(
cfw_
= = (A B) (A B)c =
[2]
(1,2,6)
(A (A B) (B (A B) = (A (A B)c ) (B (A B)c )
(10)
[.]
R Q49E4 H F E
SaPIGaD
Y
`A
W T T
XUVU
R Q49E4 H F E
SB@PIG@D
(.)
e b b
fcdc
C
4 ( B@BgBI31
A 49 46 4 2
27:
= (A B) (B Bc ) = (A B) = (A B)
(10)
[2]
[1]
(A B) B = (A B) Bc
1.
Exercise set 3.1
2b: Yes: 6r + 4s2 + 3 = 2(3r + 2s2 + 1) + 1, where 3r + 2s2 + 1 is an integer (since
r and s are integers). Therefore, it is odd, by denition.
4: Yes: for a = b = 0, we have 0 + 0 = 0 = 0 + 0.
9: The values of n2 n+11 for 1 n 10 are (i
= 414720
10
3
# !
$"
Q B 'H
US @ 9IF
'
Q B 'H
US @ 9IF
'H X V T S H
9YWU7Q ' @ '
3
i
61
i
61
g f1 c
h6e)
p
p
Ai w
S
'
9H B
'
SF
$DP
S
X
P IX B
H'
S@
T
p
1 B i qp
f
g
hf A i x9hp B Wf
w p1 g
g
Ri hf @ A 61 B h91 61
1
i
gf p
f
g
g
A i B p B hf B B ec
n
1.
Exercise set 3.2
28: Suppose c is a root of a (nonzero) polynomial with rational coecients:
P(c) =
a1
a0
an n
c + + c +
= 0,
bn
b1
b0
where ai , bi Z, bi = 0, and an = 0. We can multiply this equation by
bn b1 b0 = 0, to get:
P (c) = dn cn + + d1 c + d
Verify the following logical equivalences:
1.
(p q ) p p
2.
(p q ) (p q ) p
3.
( p q ) (p q ) p
4.
p ( q p ) p
5.
(p q ) ( p q ) q
6.
(p q ) ( p q ) c
7.
(p q ) [ p (p q )] t
8.
[( p q ) (p r)] ( p q ) (p r)
9.
(r p) [( r (p q ) (r q )] p q
10. [( p q )
EXAMPLE 1: Prove that
3 | n3 n
()
for any integer n 0.
Proof:
STEP 1: For n=0 () is true, since 3 | 03 0.
STEP 2: Suppose () is true for some n = k 0, that is 3 | k 3 k.
STEP 3: Prove that () is true for n = k + 1, that is 3 | (k + 1)3 (k + 1). We have
(k
EXAMPLE 1: Prove that
3 | 4n 1
()
for any integer n 1.
Proof:
STEP 1: For n=1 () is true, since 3 | 41 1.
STEP 2: Suppose () is true for some n = k 1, that is 3 | 4k 1.
STEP 3: Prove that () is true for n = k + 1, that is 4 | 4k+1 1. We have
4k+1 1 = 4 4k
DEFINITION: If p and q are statement variables, the conditional of q by p is If p then q
or p implies q and is denoted p q. It is false when p is true and q is false; otherwise it is
true. We call p the hypothesis of the conditional and q the conclusion.